Question

A massless rod S having length `2l` has equal point masses attached to its two ends as shown in figure. The rod is rotating about an axis passing through its centre and making angle `alpha` with the axis. The magnitude of change of momentum of rod i.e., `|(dL)/(dt)|` equals

A. `2 m l^(3) omega^(2) sin theta. cos theta`
B. `ml^(2) omega^(2) sin 2 theta`
C. `ml^(2) sin 2 theta`
D. `m^(1//2) l^(1//2) omega sin theta. Cos theta`

Answer 1

Correct Answer - B
The radius of the circular followed by the masses is `r = l sin alpha`
As, angular momentum, `L = r xx p = r xx mv`
`rArr |L| = l sin theta (m omega l sin theta)`
On differentiating, we get
`(d|L|)/(dt)=momega l^(2)2 sin theta. cos theta (d theta)/(dt)`
`rArr |(dL)/(dt)|=2ml^(2) omega^(2)sin theta. cos theta =ml^(2)omega^(2) sin 2 theta`.